The Monty Hall Problem: Help!

JamesWhitehead

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Of course he has a choice.
No. In the cases - 2 out of 3, presumably - where he knows you are wrong, he can only open the other wrong door.

It is only in the 1 case out of 3 where you are initially right that the host has a choice of wrong doors he can open.

Your new decision is effectively giving you the chance to bet that you were initially wrong. Chances are two out of three times you were! Annoying, I know, but I am now convinced. Try that table! :)

I think our brains are looking for a way to win every time but the problem is really to say how often this strategy would pay off in a series of games.
 
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INT21

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Yes, he has a choice of either goat door.

Your options stay the same.



One can play a simplified, stripped down version.

There are two doors. One has a car, one has a goat.
The host asks you to select a door, you do.

He asks if you want to change your choice.

50/50 ?

The only difference is the host can't remove a door. It's a coin toss..

I suspect if you do this alongside the above, you will come out wit the same results.

Play it often enough and it will equal itself out to 50/50.
 

gordonrutter

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Yes, he has a choice of either goat door.

Your options stay the same.



One can play a simplified, stripped down version.

There are two doors. One has a car, one has a goat.
The host asks you to select a door, you do.

He asks if you want to change your choice.

50/50 ?

The only difference is the host can't remove a door. It's a coin toss..

I suspect if you do this alongside the above, you will come out wit the same results.

Play it often enough and it will equal itself out to 50/50.
Not the same at all. Try your paper version and you’ll see the validity of swapping.
 

INT21

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No. In the cases - 2 out of 3, presumably - where he knows you are wrong, he can only open the other wrong door.

It is only in the 1 case out of 3 where you are initially right that the host has a choice of wrong doors he can open.

.
My point is that he always has a wrong door to open.
 

INT21

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Gordon,

I'll do that. But It may be a while.

I can't think of anyone who will subject themself to half an hour of this sort of thing.
 

gordonrutter

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My point is that he always has a wrong door to open.
But to start with he has two wrong possibilities and only one correct one, it’s a one in three chance 33%. By removing one of the incorrect ones it was still a one in three chance that he had chosen an incorrect door even though there are only two doors left. If he is now given a chance to change it is a 50% chance so by changing he gets it correct 17% (in total) more of the time than if he had stuck with his original answer.
 

gordonrutter

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Gordon,

I'll do that. But It may be a while.

I can't think of anyone who will subject themself to half an hour of this sort of thing.
We’ve been at it since Feb 26th 2004!
 

INT21

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Yes, there was a 1 in 3 chance. But now one of the options is gone, you only have a 1 in two chance. And you know one of them must be correct.

I suppose one could add a coin flip into the test. Heads I change, tails I don't.

Then you could play against yourself.
 

gordonrutter

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Yes, there was a 1 in 3 chance. But now one of the options is gone, you only have a 1 in two chance. And you know one of them must be correct.

I suppose one could add a coin flip into the test. Heads I change, tails I don't.

Then you could play against yourself.
Yes, read the rest of what I said. And no, no coin flip necessary if you do you reduce your overall chance of coming out ahead.
 

EnolaGaia

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Much of the confusion underlying debates about the Monty Hall Problem relates to the implied context of the game show scenario.

There was no standard routine or protocol for revealing the final Big Deal during the early days of the game show (1960s).

When Marilyn vos Savant wrote her 1990 response to a reader's question she failed to specify all the presumptions underlying her answer. She then received a wave of rebuttals, many of which rightly pointed out that her solution entailed certain presumptions about the randomness of certain selection opportunities, the host's actions, the possibility of switching prizes behind the doors during the first reveal, etc.

Von Savant saved face by eventually clarifying what presumptions were entailed in her answer, and her answer was sound in the context of these subsequently specified presumptions. Whether or not these later-described presumptions were built into, and / or adequately communicated, in her original 1990 answer is itself a matter of debate.

However ... Unless one recognizes and accepts the same set of presumptions, one can just as reasonably assign different probabilities to the same basic scenario.
 

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sorry, I simply don't agree.

You say above 'if the car/goat were shuffle once one door was opened'. I don't see any reference to the doors being shuffled. even if they were, it wouldn't matter as there would still be only two doors, and so it is still the same choice; stay or move.

Let's assume you picked the correct door. Monty opens a goat door (as he is always going to do as he is simply removing a losing door from the equation).

So, you have picked the correct door. Why should moving improve your chances, you have already picked the winner.

Also, why four pieces of paper ? there are only three doors. surely you mean three pieces, one marked.

The way I see it, he is playing two games. In the first one, your odds are 1 to 3.

Then there is a new game in which your odds are 1 to 2. The result of your choice in the first game has no effect on the new game.

Anyway, I will try the idea of using paper. I had considered it already.

Maybe play it one hundred times. Fifty I don't change, fifty I do. I'll get back to you with the results. May be a few ays though. Rather busy at the moment.

INT21.
Sorry, I wrote my post very quickly and hit 4 instead of 3. The prizes are never shuffled behind the doors, I was just saying that if that happened then yiour choice would revert to 50/50.

I even talk about this now in one of my shows and I tried to explain it with a new angle.

You know the car is behind one of the 3 doors. Therefore your first choice has a 1 in 3 chance of being correct. The host then opens a losing door. There are now two remaining closed doors.

If a new person/player was now asked to choose a door, then they have a 50/50 chance. You however, have more information, as you made a choice from 3 doors, not from 2. So when the host asks if you want to swap, you are not making a new choice based on 2 doors, you are revising your first choice based on 3 doors. And probability says that your first choice was incorrect. Therefore the car is somewhere else.

There is also an online version you can do here: https://www.mathwarehouse.com/monty-hall-simulation-online/

I just did it 30 times, swapped every time and got 63% cars and 37% goats.
Anyway, try it yourself and see what happens.
 

INT21

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Back in post 33, Min Bannister wrote.

''I thought this thread would die after two posts as well!''

Ah well. didn't work out that way.
 

INT21

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Sorry, I wrote my post very quickly and hit 4 instead of 3. The prizes are never shuffled behind the doors, I was just saying that if that happened then yiour choice would revert to 50/50.

I even talk about this now in one of my shows and I tried to explain it with a new angle.

You know the car is behind one of the 3 doors. Therefore your first choice has a 1 in 3 chance of being correct. The host then opens a losing door. There are now two remaining closed doors.

If a new person/player was now asked to choose a door, then they have a 50/50 chance. You however, have more information, as you made a choice from 3 doors, not from 2. So when the host asks if you want to swap, you are not making a new choice based on 2 doors, you are revising your first choice based on 3 doors. And probability says that your first choice was incorrect. Therefore the car is somewhere else.

There is also an online version you can do here: https://www.mathwarehouse.com/monty-hall-simulation-online/

I just did it 30 times, swapped every time and got 63% cars and 37% goats.
Anyway, try it yourself and see what happens.
In a way, this is similar to the problem of the three friends and the bent waiter. But that relied upon the friends not knowing that the cost of the meal had changed.

But here, there is really no mystery. You know what has changed.

You don't have any more information than a new player coming in after the first door is opened would have.

He would be presented with a choice of two doors. One has the car.

You know the same. Two doors. One has the car.

If a new player entered after the goat door is eliminated the hosts question would be ' There is a car behind one of these two doors. which one do you pic ?

You pick one. The odds are 1 in 2.

If the host says ' do you want to change your mind ? Then the odds are still one 1 in 2.

You will either get the car or the goat.

I'm definitely going to try this with cards.


INT21.

p.s. Re the removed post. well, I thought it was funny ;)

Hint taken.
 
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INT21

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I thought this thread would die after two posts as well!

OKay lets try it this way:

There are two goats and only one car.

Therefore if you pick a door you have more likely chosen a goat.

The host shows you another goat behind one of the other doors. So you know that there is a goat behind that door. And you also know that you have probably picked a goat with your first choice because that is where the odds lie.

The best option is therefore to switch as the car is then most likely to be behind the other door. :)

Fabulous book by the way.:yeay:
@Min Bannister ,

Why do you know you have probably picked a goat ? He is going to show you a goat whatever you pick ?
 

INT21

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An aside.

Is it true there are only six ways it is possible to present the doors ?

a,b,c
b,c,a
c,a,b
b,a,c
c,b,a
a,c,b

If so I need two random number generators (or use one twice) and a coin to do it properly.

And three series of 100 readings.
 
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INT21

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So, here we go. The grand reveal.

Equipment.

I butter tub with 3 * 20mm polystyrene spheres marked A,B and C. (Random number generator)

1 butter tub with 2 * 20 mm spheres marked Y and N (second random number generator)

Note pad marked of in columns. Column headings Run, Door, Pick, Change, Win/Lose

Two sets of runs carried out. 100 runs in each set. First set was just to establish the random nature of the results and no change was made to the selected door choice. So the yes (Y)/No(N) balls were not used.

I am right handed, so the tubs were positioned to my left so I could not easily see them while I recorded the results on the pad to my right.

Procedure. starting with run 1

Select a ball. write down the result in 'DOOR' column. drop ball back in tub.
Select a ball. write result in 'PICK' column. drop ball back in tub.

Does first ball match second ball ? If yes write W for win in .WIN/LOSE column, L for lose if they don't match.

Note, the 'Change' column not needed for this test.

Carry out the remaining 99 runs.

Results.

68 Loses
32 Wins.

so one can assume that if the change option had been used every time then the results would have been reversed. I.e 68 wins to 32 Loses.

Test two.

Using the two Y/N balls as a coin toss. This simulates whether you change your mind or not.

Note that it is assumed the host will remove a goat door every time. So you only have two choices.

Procedure.

Select ball from a,b,c tub. Note letter and place in DOOR column. replace ball in tub.

Select ball from a,b,c tub. Note letter and place in CHOICE column. Replace ball in tub

Select a ball from Y/N tub. If Y, then if you had picked the correct door, it becomes the incorrect door, and you lose. Likewise if you picked the incorrect door it becomes the correct one and you win.

Write W or L in WIN/LOSE column.

Do this for the other 99 runs.

Results.

41 wins.

On 16 occasions changing the choice resulted in a lose from a win.

On 16 occasions not changing resulted in a win from the original choice.

But basically changing the choice only resulted in 41% win.

INT21.
 

INT21

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That is for one set of results. Maybe I'll do it again when I have devised a computer program for it.

There is something I wish to try.

Can one assume that the host changes the car door between each contestant ?

It would make a difference as, if not, you are using only one set of variables, not two as I have done. One set for which door the car is behind, and one set for the door I pick.
 

Mikefule

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I was perplexed by the Monty Hall problem until a friend (a Professor who specialises in probabilities and statistics) suggested I analysed with a simple tree diagram such as we did in O Level (now GCSE) maths many years ago. Then it was obvious. It is counterintuitive, but definitely the case that changing is the best option.

When the player chooses a door, they have no information except that there is a good thing (e.g. a car) behind one door, and a bad thing (e.g. a goat) behind each of the other two doors. (Not that goats are all bad, of course.)

Therefore, the player is making a purely random choice. There is a 1/3 chance they have chosen the car, and a 2/3 chance they have chosen a goat.

The crucial thing is that the host (Monty) KNOWS what s behind each door.

Monty will only open a door that does not have a car behind it, otherwise that would give the game away.

Therefore, Monty's action gives further information to the player.

He has not changed the odds ont he player's first guess having been right, because nothing has changed about that door. It is still 1/3.

The other 2 doors, as a pair, were worth a 2/3 chance of the car being there. Monty eliminates one of those doors and in effect therefore says "There is a 2/3 chance that the car is behind one of these two doors, and it certainly isn't this one, so it must be either the other one of this pair, or the one you've already chosen."

So the player has a 1/3 chance if they stick, and a 2/3 chance if they switch.

The key thing is that Monty's action is not random; it is based on certain knowledge, and therefore he is giving new information to the player.

Absolutely counterintuitive to most of us, but once you draw a diagram it makes sense.
 

INT21

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However the player only has one option at that stage. Stick or change. And he knows the car is behind one if the two doors.

He had a one in three chance in the first part of the game of picking the right door' and he may have already done so. Now it is down to one in two. Monty was going to remove a goat door anyway.
if the player dropped dead and someone else came in at that stage and took his place, the new players choice would be 50/50.
 

INT21

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Thinking more about this.

Monty's knowledge doesn't effect anything.

He is going to remove a goat door otherwise there is no point in the game. He can't remove the car door. And the player knows this.

so he removes a goat door and both he and the player have the same knowledge. That being that the car is behind one of the two remaining doors; it has to be.
Monty knowing which one is not going to effect the game. The only thing left to do is for the player to change or stick.

Monty has no say in this decision.
 

INT21

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This is coming across as one of those mathematical sophistries where two plus two can be made to equal five.
Mathematical slight of hand.
 

JamesWhitehead

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The whole problem hinges on the confusion between the fifty-fifty choice of doors, which is true in any given iteration of the game and the odds of switching against sticking, which is a different question. Took me a while to get that! :)
 

INT21

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On 14 of the runs I picked the correct door then swapped out and lost.
 

INT21

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Just tried something else.

If I had swapped every time I would have won 52% of the time.

Not swapping at all, 32%
 

INT21

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Gordon,

Agreed.

But I will need to run the thing, say, ten times to feel comfortable with the premise that changing will ALWAYS result in better results than not changing.

I get around to writing a program based on the current manual results. when I can duplicate that I will run more tests.

It does appear, based on the one set of figures that it is true.
 

oxo66

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On 16 of the runs I picked the correct door then swapped out and lost.
Blessmycottonsocks said:
"changing the choice only resulted in 41% win." Which is near as damn it mid way between 33% and 50%.
Couldn't be more inconclusive if you tried!
This is because you introduced the extra random element in your trial:

For your baseline test 1 you never swapped. This the usual default, but
for your alternative test 2 you had 50:50 swap or no swap. The alternative should be ALWAYS swap.
(There shouldn't be a 50:50 choice at this stage because Monty ensures that your loser will become a winner - if you swap!)

Your test 2 should have had about 50 swaps and 50 no-swaps (?)
The no swaps would give you about 17 wins from 50 (you said 16, fine), and
The swaps should give you about 33 wins from the other 50 (you implied 41-16 = 25 wins which does seem low and suspiciously close to half of 50...)

Anyway message is, if you do do a test again take out the 50:50 bit.
Having seen the exchange since I started writing this, this may not help your understanding. But it helps mine.

O
 

Bad Bungle

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This is coming across as one of those mathematical sophistries where two plus two can be made to equal five.
Mathematical slight of hand.
No disrespect to anyone but there is a danger of over-thinking the problem. This isn't a Fortean exercise in predicting the future or X ray vision or whether the quantum state of an object isn't fixed until it's door is opened, it's just mechanical non-sentient mathmatical probabilities that have no memory of past events. If you don't like or understand the Monty Hall aspect to it then ignore it as Marilyn vos Savant did.

Three doors, one car, two goats: pick a door and the odd/probabilities/chances are 1/3 that there is a car behind it and 2/3 that there isn't.
And that's about it.
Take your polystyrene spheres or rubber ducks or pieces of paper or whatever method you have for a random selection, choose a "door" and then reveal the other two choices. Repeat thirty times and you should get (roughly) 10 wins and 20 losses
For added tension you can choose a door and instead of revealing both of the other doors simultaneously, you can open one and then insert a faux-dramatic pause and say "is that your final answer?" in a nasal tone before revealing the second one. It makes no difference to the odds, repeat thirty times and the probabilities are still that the car is behind your door on 10 occasions and the car is behind one of the other doors on 20 occasions.
What the Host (Monty) does after you picked a door was reveal one of the other doors and then pause. Marilyn vos Savant didn't say whether Monty knew where the car was, she just said that the Host opened one of the remaining doors and it was a goat.
The probabilities haven't changed that when you chose your door, the car was twice as likely to be behind another door. Monty has just shown which of the two other doors it wasn't behind (I believe this has something to do with "reducing the degree of freedom") so ... the greater odds are the remaining door has the car - so switch.

What is interesting to me is that apparently in real life Game Show scenarios, so few Contestants switch when given the choice.
 

INT21

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oxo66, Bad Bungle,

I have more or less resigned to the fact that it appears to happen. I just don't follow the reasoning.

However, as I stated above, I'll do another ten runs to see if the average is above 50% at the end.

But not for a while. It took 40 minutes to do the last one.
 
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